Question 1 of 10
Which object has the most gravitational potential energy?

A. A 8 kg book at a height of 2 m
B. An 8 kg book at a height of 3 m
C. An 5 kg book at a height of 3 m
D. A5 kg book at a height of 2 m
SUBMIT

Answer:

c

Explanation:

Answer:

Density is affected by volume and mass.

Explanation:

Density is defined as the quantity of mass per unit of volume, or expressed mathematically, d = m/v.

Answer:

one sec let me think

Explanation:

(a)The average velocity of the ball over the following time intervals will be  [3,4] ft/sec.

(b)The instantaneous velocity at time t = 3 will be32 ft/sec.

(c)The instantaneous velocity at time t = 6 will be -64 ft/sec.

(d)The ball will hit the ground at 13.4 sec.

The change of displacement with respect to time is defined as the velocity.  velocity is a vector quantity. it is a time-based component.

The given data in the question will be ,

u is the initial velocity by which ball thrown=128 ft/sec.

V₃ is the instantaneous velocity at time t=3 sec.

V₆ is the instantaneous velocity at time t=6 sec.

t is the time when ball hits the ground=?

(a) Given equation for the displacement

s(t) =128t-16t²     (on differenting got the velocity )

v(t) = 128-32t

Time when velocity is zero will be

[tex]\rm{ t=\frac{128}{32}[/tex]

[tex]\rm{ t=4 sec[/tex]

If the velocity got in the equation is 128 and 32 ft /sec. it can be only when the average velocity is [3,4] ft/sec .

Hence the average velocity obtained from the problem will be  [3,4] ft/sec

(b)  

s(t) =128t-16t²     (on differenting got the velocity )

v(t) = 128-32t

At time( t=3 sec)

v(3) = 128-32×3

v(3) =32 m/sec.

Hence the instantaneous velocity at time t = 3 will be32 ft/sec.

(c)

s(t) =128t-16t²     (on differenting got the velocity )

v(t) = 128-32t

At time( t=6 sec)

v(6) = 128-32×6

v(6) = -64 m/sec.

Hence the instantaneous velocity at time t = 6 will be -64 ft/sec.

(d)

According to Newtons third equation of motion we got

v=u+gt

If the body returens from a certain height at max height its velocity must be zero; ( u=0)

[tex]\rm t=\frac{(v-u)}{g} \\\\\ \rm t=\frac{(128-0)}{9.81}\\\\\rm t=13.04 sec.[/tex]

Hence the ball will hit the ground at 13.4 sec.

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When an object is raised vertically, work is done against _gravitational force _________

The force of attraction between all masses in the universe; especially the attraction of the earth's mass for bodies near its surface is called gravitational force .

On every body there is a force acting named  gravitational force which attracts the body downward .

If a object / body is thrown up it always comes downward if only gravitational force is acting on it which act due to earth's gravity .

When an object is raised vertically, work is done against _gravitational force _________

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Answer:

a)       [tex]- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o} ) \frac{sin \theta}{R }[/tex] =  [tex]\frac{d^2 \theta}{d t^2}[/tex]

b)     f = 2π  [tex]\sqrt{ \frac{R}{ g - \frac{q}{m} \frac{\sigma }{2 \epsilon_o} } }[/tex]  

Explanation:

a) To have the equations of motion, let's use Newton's second law.

Let's set a reference system where the x-axis is parallel to the path and the y-axis is in the direction of tension of the rope.

For this reference system the tension is in the direction of the y axis, we must decompose the weight and the electrical force.

Let's use trigonometry for the weight that is in the vertical direction down

             sin θ = Wₓ / W

             cos θ = W_y / w

             Wₓ = W sin θ

             W_y = W cos θ

we repeat for the electric force that is vertical upwards

              F_{ex} = F_e sin θ

              F_{ey} = F_e cos θ

the electric force is

               F_e = q E

where the field created by an infinite plate is

               E = [tex]\frac{ \sigma}{2 \epsilon_o}[/tex]  

let's write Newton's second law

Y  axis  

           T - W_y = 0

            T = W cos θ

X axis

            F_{ex} - Wₓ = m a                   (1)

we use that the acceleration is related to the position

            a = dv / dt

            v = dx / dt

where x is the displacement in the arc of the curve

substituting

            a = d² x /dt²

we substitute in 1

           q E sin θ - mg sin θ = m [tex]\frac{d^2 x}{dt^2}[/tex]

we have angular (tea) and linear (x) variables, if we remember that angles must be measured in radians

           θ = x / R

           x = R θ

we substitute

           sin θ (q E - mg) = m \frac{d^2 R \ theta}{dt^2}  

          [tex]- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o} ) \frac{sin \theta}{R }[/tex] =  [tex]\frac{d^2 \theta}{d t^2}[/tex]

this is the equation of motion of the system

b) for small oscillations

         sin θ = θ

therefore the solution is simple harmonic

      θ = θ₀ cos (wt + Ф)

if derived twice, we substitute

- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o}  ) \frac{\theta}{R } θ₀ cos (wt + Ф) = -w² θ₀ cos (wt + Ф)

       w² =  [tex]\frac{g}{R}[/tex] - [tex]\frac{q}{m} \frac{ \sigma }{2 \epsilon_o} \frac{1}{R}[/tex]  

angular velocity is related to frequency

       w = 2π f

        f = 2π / w

        f = 2π/w

        f = 2π  [tex]\sqrt{ \frac{R}{ g - \frac{q}{m} \frac{\sigma }{2 \epsilon_o} } }[/tex]  

Answer:

It's b because he discovered galaxies .

Answer:

A wave can be thought of as a disturbance or oscillation that travels through space-time, accompanied by a transfer of energy. The direction a wave propagates is perpendicular to the direction it oscillates for transverse waves. A wave does not move mass in the direction of propagation; it transfers energy.

Answer:

Examples of compound machines include bicycles, cars, scissors, and fishing rods with reels. Compound machines generally have lower efficiency but greater mechanical advantage than simple machines

Brilianst

Answer:

screws, inclined planes , etc

Explanation:

no explanation needed

Answer:

Thermal Energy

Explanation:

The energies that work together to bake the bread and cook the eggs is the thermal energy. Thermal energy is basically heat energy, which makes the food warmer.

Thermal energy and chemical energy work together to bake the bread and cook the eggs.

The energy present in a system that determines its temperature is referred to as thermal energy. Thermal energy flows as heat. Thermodynamics is a whole field of physics that studies how heat is transmitted across various systems and how work is done in the process.

Chemical energy is described as: the power that is kept in chemical compound bonds (molecules and atoms). It is released during the chemical process, which is referred to as an exothermic reaction, which mostly generates heat as a byproduct.

When the bread is baked or the eggs are cooked, thermal energy is provided from outside and  chemical energy inside the molecules of bread  or eggs come to play. So, these two energies work together to bake the bread and cook the eggs.

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Time = (20 km) x (1 hr/2 km) x (3,600 sec/hr)

Time = (20 x 1 x 3,600 km-hr-sec) / (2 km-hr)

Time = (20 x 1 x 3,600 / 2) (km-hr-sec / km-hr)

Time = 36,000 seconds

(That's 10 hours.)

[tex]\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}[/tex]

Actually Welcome to the Concept of the Waves.

Wavelength is the distance between two consecutive crest or trough.

hence, here the distance is 10cm

So the wavelength is 10cm

===> 10 cm

Answer:

24.5 m/s

Explanation:

KE=1/2mv^2

15000=1/2(50)v^2

30000=(50)v^2

600=v^2

sqrt600=v

v=24.5 m/s!!

Answer: The main difference between the three is the mode of transmission. The chest pass is straight through the air towards your teammate. While the bounce pass is directed toward the ground and then at your teammate. Finally, the overhead pass is projected high in the air to avoid defenders.

Explanation:

Answer:

The main difference is: the chest pass is straight through the air towards your teammate. While the bounce pass is directed toward the ground and then at your teammate.

The secondary main difference is the amount of power from the ball recived from the person reciving

Answer:

d

Explanation:

The uneven heating of land and water in coastal areas creates fronts that cause thunderstorms. So, option D is correct.

Heat energy of a system is defined as the energy transferred between the molecules due to the temperature difference between them. Heat energy is the energy responsible for the temperature of the molecules.

Here,

A city near the ocean experiences thunderstorms on a frequent basis. This is because,

The atmosphere near to the land is highly humid and warm and thus creates a relatively low pressure in the land. In order to balance the pressure variation the air from the oceans make movements towards the land, and as a result, sea breeze are formed. During the evenings when the water in the ocean is more warm and when land begins to cool, the air from land rushes towards the sea, in order to balance the pressure variation occurred here. This makes the air above the ocean highly humid and thus would lead to the formation of thunderstorms.

Hence,

The uneven heating of land and water in coastal areas creates fronts that cause thunderstorms.

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Answer:

B is the Answer

Follow me please

Mark brainliest

Answer:

The answer is A

Explanation:

This is because f = 1/T,

or frequency equals 1 over period, T

Since the highest T would be the answer with the greatest length and mass does not matter the answer would be c IF they were asking for the highest period.

F is found by 1 over T so the greatest F would be the smallest number, Hence answer A

Please Mark me brainiest

Answer:

rotates

Explanation:

I'm so bored

yrfgggghhgghhyuj

Explanation:

oh no i hope i dont fall and the  i agree with the other dude

Answer: Okay, thanks I really appreciate the generosity.

Answer:

Lost Mechanical Power = 7.7565 KW

Head Loss = 26.35 m

Explanation:

First, we will find the useful mechanical power used to transport water to the higher reservoir:

[tex]P_{useful} = \rho ghV[/tex]

where,

P_useful = Useful mechanical Power = ?

ρ = density of water = 1000 kg/m³

g = acceleration due to gravity = 9.81 m/s²

h = height = 45 m

V = Volume flow rate = 0.03 m³/s

Therefore,

[tex]P_{useful} = (1000\ kg/m^3)(9.81\ m/s^2)(45\ m)(0.03\ m^3/s)\\P_{useful} = 13243.5\ W = 13.2435\ KW[/tex]

Now, the lost mechanical power will be:

[tex]Lost\ Mechanical\ Power = Total\ Mechanical\ Power - Useful\ power\\Lost\ Mechanical\ Power = 21\ KW - 13.2435\ KW\\[/tex]

Lost Mechanical Power = 7.7565 KW

Now, for the head loss:

[tex]Lost\ Mechanical\ Power = \rho g(Head\ Loss)V\\Head\ Loss = \frac{Lost\ Mechanical\ Power}{\rho gV} \\\\Head\ Loss = \frac{7756.5\ W}{(1000\ kg/m^3)(9.81\ m/s^2)(0.03\ m^3/s)} \\[/tex]

Head Loss = 26.35 m

Answer:

a. 20 s

b. 0 m/s  

c. right

d.no its inelastic because when the car b was at rest and a was coming in at it, since b had no force what so ever car a swept it away with it moving to the right

Explanation:

im not sure though

By applying conservation of linear momentum, the answers are:

1. Time = 2 s

2. 3 m/s

3. same direction

4. Inelastic collision

There are for types of collision. They are;

Given that a local church is hosting a carnival which includes a bumper car ride. Bumper car A and its driver have a mass of 300 kg; bumper car B and its driver have a mass of 200 kg. Bumper car A has a velocity to the right of 2 m/s and bumper car B is at rest. At t = 0 s, bumper car A and B are separated by 10 m. Bumper car A accelerates at 1 m/s2 to a velocity of 4 m/s and continues at this constant speed until colliding with bumper car B.

1. The time required for bumper car A to travel the 10 m to collide with bumper car B can be calculated by using first equation of linear motion.

V = U + at

Where

V = 4 m/s

U = 2 m/s

a = 1 m/[tex]s^{2}[/tex]

Substitute all the parameters into the formula

4 = 2 + t

t = 4 - 2

t = 2s

2. To calculate the speed of bumper car A following the collision with bumper car B, which now has a velocity to the right of 3 m/s, we will apply conservation of linear momentum

[tex]m_{1}u_{1}[/tex] = [tex]m_{1}v_{1}[/tex] + [tex]m_{2}v_{2}[/tex]

300 x 4 = 300V + 200 x 3

1200 = 300V + 300

300V = 1200 - 300

300V = 900

V = 900/300

V = 3 m/s

3. Since the final velocity of car A is positive, the direction of motion for bumper car A follows the collision with bumper car B to the right.

4. Since the both move at the same velocity, the collision inelastic.

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Answer:

a = 1,538 m / s²

Explanation:

Let's use Newton's second law, let's set a reference system where the x-axis is parallel to the sloping floor of the truck and the positive direction is in the direction of movement of the trough, for this case the weight is the only force to decompose

          sin θ = Wₓx / W

          cos θ = W_y / W

          Wₓ = W sin θ

          W_y = W cos θ

Y axis

          N -W_y = 0

          N = mg cos θ

X axis

          Wₓ - fr = m a

the friction force has the expression

          fr = μ N

There are values ​​of the friction coefficient (μ_s) one for when the movement has not started and it takes a smaller value for when the bodies are moving.

In this case we first find the angle for which the movement begins, in this part we use the static coefficient and the acceleration is zero

             Wₓ - μ_s N = 0

             m g sin θ = μ_s mg cos θ

             tan θ = μ_s

             θ = tan⁻¹ μ_s

we calculate

              θ = tan⁻¹ 0.8

              θ = 38.7º

For this angle, how the trough begins to move, the coefficient is reduced to the dynamics coefficient (μ_k)  and the acceleration is different from zero.

we substitute

          mg sin θ - μ_k mg cos θ = m a

          a = g (sin θ - μ_k cos θ)

let's calculate

          a = 9.8 (sin 38.7 - 0.6 cos 38.7)

          a = 1,538 m / s²

Answer:

Me encanta el pollo, de todos modos, un día que es hoy, por cierto, fui a la tierra de los pollos y caminé en México, ¿sabías que es mi sueño? ¿Olvidé que mi nombre es jemma y me gusta caminar? párrafo así que nos vemos

Explanation:

Answer:

Fluorine needs to gain only 1 electron to have a full octet, while nitrogen needs to gain 3.

Answer:

t_pass = 2.34 m

t_stop = 4.68 s

Thus, for the car passing at constant speed the pedestrian will have to wait less.

Explanation:

If the car is moving with constant speed, then the time taken by it will be given as:

[tex]t_{pass} = \frac{D}{v}[/tex]

where,

t_pass = time taken = ?

D = Distance covered = 23 m

v = constant speed = (22 mi/h)(1609.34 m/1 mi)(1 h/3600 s) = 9.84 m/s

Therefore,

[tex]t_{pass} = \frac{23\ m}{9.84\ m/s} \\[/tex]

t_pass = 2.34 m

Now, for the time to stop the car, we will use third equation of motion to get the acceleration first:

[tex]2as = v_{f}^{2} - v_{i}^2\\a = \frac{v_{f}^{2} - v_{i}^2}{2D}\\\\a = \frac{(0\ m/s)^{2}-(9.84\ m/s)^2}{(2)(23\ m)}\\\\a = -2.1\ m/s^2[/tex]

Now, for the passing time we use first equation of motion:

[tex]v_{f} = v_{i} + at_{stop}\\t_{stop} = \frac{v_{f}-v_{i}}{a}\\\\t_{stop} = \frac{0\ m/s - 9.84\ m/s}{-2.1\ m/s^2}[/tex]

t_stop = 4.68 s

Constant velocity is the velocity which covers the same distance for each interval of the time.

The time required to pass is 2.34 seconds and the time to stop is 4.68 seconds.

Constant velocity is the velocity which covers the same distance for each interval of the time.

It can be given as,

[tex]v=\dfrac{x}{t}[/tex]

As the distance covered by the car is 23 meters and the constant velocity of the car is 22 miles per second.

Convert the unit of velocity in m/s the value obtained will be 9.84 m/s.

Thus amount of time, [tex]t_{pass}[/tex] is,

[tex]9.84=\dfrac{23}{t_{pass} } \\t_{pass} =\dfrac{23}{9.84} \\t_{pass} =2.34[/tex]

As the distance covered by the car is 23 meters and the constant velocity of the car is 22 miles per second.

Convert the unit of velocity in m/s the value obtained will be 9.84 m/s.

Thus amount of time, [tex]t_{pass}[/tex] is,

[tex]9.84=\dfrac{23}{t_{pass} } \\t_{pass} =\dfrac{23}{9.84} \\t_{pass} =2.34[/tex]

According to the third equation of the motion acceleration can be given as,

[tex]v^2-u^2=2ax\\a=\dfrac{v^2-u^2}{2x}\\a=\dfrac{0^2-9.84^2}{2\times 23}\\a=-2.1 \rm \; m/s^2[/tex]

Now, use the first equation of motion, to get the required time,

[tex]v=u+at\\0=9.84+(-2.1)t\\t=4.68\rm \; s[/tex]

Therefore, the time required to pass is 2.34 seconds and the time to stop is 4.68 seconds.

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Answer:

1237 J

Explanation:

The work done by the hose on the balloon is the work done by a spring which is

W = 1/2k(x₀² - x₁²) where k = spring constant = 112 N/m, x₀ = 4.70 m and x₁ = 0 m.

Substituting the values of the variables into the equation, we have

W = 1/2k(x₀² - x₁²)

W = 1/2 × 112 N/m((4.70 m)² - (0 m)²)

W = 56 N/m(22.09 m² - 0 m²)

W = 56 N/m(22.09 m²)

W = 1237.04 J

W ≅ 1237 J

Answer:

yes

Explanation: